D1 HELP
Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.
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D1 HELPQuestion 5c, how do you do it ? Can someone explain please, link to the paper is below, if the link is broken, the paper is January 2011.
http://www.edexcel.com/migrationdocu...e_20110119.pdf -
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Re: D1 HELPYou must first make the graph semi-Eulerian as the question states that he can start at D and finish at any other vertex.
So let's list all the odd valencies and the possibilities that you can go through (which should have been done for part a but I'll list them here anyway).
AD + FI = 4.5 + 5.3 = 9.8
AF + DI = 5.8 + 3.9 = 9.7
AI + DF = 5.9 + 5.1 = 11
He must start at D so we must eliminate any paths that include D. This means that we have to choose between FI, AF, or AI. We want the route to be minimised so we want to choose the route that gives the shortest length.
FI gives the shortest length (5.3) so therefore he must repeat FI. This means that he starts at D and ends at A.
Hopefully that was a clear explanation, but I can clarify anything if needed. -
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Re: D1 HELPI do not understand how you came to those conclusions. Elaborate please ?(Original post by MrJames16)
You must first make the graph semi-Eulerian as the question states that he can start at D and finish at any other vertex.
So let's list all the odd valencies and the possibilities that you can go through (which should have been done for part a but I'll list them here anyway).
AD + FI = 4.5 + 5.3 = 9.8
AF + DI = 5.8 + 3.9 = 9.7
AI + DF = 5.9 + 5.1 = 11
He must start at D so we must eliminate any paths that include D. This means that we have to choose between FI, AF, or AI. We want the route to be minimised so we want to choose the route that gives the shortest length.
FI gives the shortest length (5.3) so therefore he must repeat FI. This means that he starts at D and ends at A.
Hopefully that was a clear explanation, but I can clarify anything if needed.
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Re: D1 HELPOk, so the question states that we must start at D and end at any other vertex.(Original post by iAre Teh Lejend)
I do not understand how you came to those conclusions. Elaborate please ?
By making a graph semi-Eulerian, it means that we are able to traverse each road at least once by starting at an odd valency and finish at the other odd valency. A, D, F and I are the odd valencies in this question. So we have to alter the graph so that there are only two odd valencies.
So we must start at D, and we must ensure that it is kept as an odd valency so that we can start from D. So we have to consider the other valencies: A, F and I.
As I stated earlier, FI is the shortest route so we must repeat FI (or it should be FHI, rather). Note that this makes F and I even valencies, and keeps D and A as odd valencies, so we can start at D and end another vertex, A. -
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Re: D1 HELPAhhh so basically, we can only start and end at odd valencies ?(Original post by MrJames16)
Ok, so the question states that we must start at D and end at any other vertex.
By making a graph semi-Eulerian, it means that we are able to traverse each road at least once by starting at an odd valency and finish at the other odd valency. A, D, F and I are the odd valencies in this question. So we have to alter the graph so that there are only two odd valencies.
So we must start at D, and we must ensure that it is kept as an odd valency so that we can start from D. So we have to consider the other valencies: A, F and I.
As I stated earlier, FI is the shortest route so we must repeat FI (or it should be FHI, rather). Note that this makes F and I even valencies, and keeps D and A as odd valencies, so we can start at D and end another vertex, A. -
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Re: D1 HELPYes, if you want to finish at another vertex (which must be an odd valency).(Original post by iAre Teh Lejend)
Ahhh so basically, we can only start and end at odd valencies ?
If you try to devise a route for this new graph then you will find that when you start at D you will finish at A, and you will only repeat FHI.
